The curves \(C_1\) and \(C_2\) are defined by \[y=e^{-x}\qquad (x>0)\qquad \text{and}\qquad y=e^{-x}\sin x \qquad (x>0),\] respectively. Sketch roughly \(C_1\) and \(C_2\) on the same diagram.

Graph of y = e to the power of minus x sin x (C2), with the graphs of y equals e to the minus x (C1) and minus e to the minus x both marked. These two decay exponentially from one and minus one respectively, towards zero, while the graph of e to the minus x sin x oscillates between them, touching each of the two curves at regular intervals.

For \(C_1\) and \(C_2\), we need the parts of these curves that satisfy \(x > 0\).

The two curves \(\,y=e^{-x}\) and \(\,y=-e^{-x}\) form what is known as an `envelope’ around \(\,y=e^{-x}\sin x\).

Let \(x_n\) denote the \(x\)-coordinate of the \(n\)th point of contact between the two curves, where \(0<x_1< x_2<\dotsb\), and let \(A_n\) denote the area of the region enclosed by the two curves between \(x_n\) and \(x_{n+1}\). Show that \[A_n=\frac{1}{2}\left(e^{2\pi}-1\right)e^{-(4n+1)\pi/2}\]

First we need to know what the points \(x_n\) are. Solving \[e^{-x}=e^{-x}\sin x,\qquad \left(e^{-x}\neq 0 \right)\] \[\Longrightarrow \sin x = 1 \qquad \Longrightarrow \qquad x_1=\frac{\pi}{2},\, x_2=2\pi+\frac{\pi}{2} \dotsb\] Thus \[x_n=2\pi(n-1)+\frac{\pi}{2}=\frac{1}{2}(4n-3)\pi\] and \[A_n=\int_{\frac{1}{2}(4n-3)\pi}^{\frac{1}{2}(4n+1)\pi}\,e^{-x}\left(1-\sin x\right)\,dx,\] the difference between the two curves \(C_1\) and \(C_2\). This can be seen below.

The same graph as before, but with the successive pieces of the space between C1 and C2 marked on as A1, A2, A3, separated by the points where C1 and C2 touch.
Consider \(I=\int e^{-x}\sin x\,dx\). We can solve this via integration by parts, with \(u=\sin x\) and \(v'=e^{-x}\): \[\begin{align*} I &= \int e^{-x}\sin x \,dx \\ &= \left[-e^{-x}\sin x \right] + \int e^{-x}\cos x \,dx \\ &= \left[-e^{-x}\sin x \right] + \left[-e^{-x}\cos x \right] - \int e^{-x}\sin x \,dx \, \, \text{ (by integrating the integral above by parts again)}\\ &= \left[-e^{-x}\sin x \right] + \left[-e^{-x}\cos x \right] - I\\ \Longrightarrow \quad 2I &= -e^{-x}\left(\sin x+ \cos x\right)\\ \Longrightarrow \quad I &= -\frac{1}{2}e^{-x}\left(\sin x+ \cos x\right) \end{align*}\] Thus \[\begin{align*} A_n &=\int_{x_n}^{x_{n+1}}\,e^{-x}\left(1-\sin x\right)\,dx \\ &= \int_{\frac{1}{2}(4n-3)\pi}^{\frac{1}{2}(4n+1)\pi} \,e^{-x}\,dx - \int_{\frac{1}{2}(4n-3)\pi}^{\frac{1}{2}(4n+1)\pi} \,e^{-x}\sin x\,dx \\ &= \left[-e^{-x} + \frac{1}{2}e^{-x}\left(\sin x+ \cos x\right)\right]_{\frac{1}{2}(4n-3)\pi}^{\frac{1}{2}(4n+1)\pi} \, \, \text{(recognising the second integral above as $I$)}\\ &= \left[\frac{1}{2}e^{-x}\left(\sin x+ \cos x-2\right)\right]_{\frac{1}{2}(4n-3)\pi}^{\frac{1}{2}(4n+1)\pi} \\ &= \frac{1}{2}e^{-\frac{1}{2}(4n+1)\pi}\left[\sin \left(2n\pi+\frac{\pi}{2}\right) + \cos \left(2n\pi+\frac{\pi}{2}\right) -2\right] \\ &\quad - \frac{1}{2}e^{-\frac{1}{2}(4n+1)\pi}e^{2\pi}\left[\sin \left(2(n-1)\pi+\frac{\pi}{2}\right) + \cos \left(2(n-1)\pi+\frac{\pi}{2}\right) -2\right]. \end{align*}\] With \(\,\sin\left(2n\pi + \frac{\pi}{2}\right)=\sin\frac{\pi}{2}=1\) and \(\,\cos\left(2n\pi + \frac{\pi}{2}\right)=\cos\frac{\pi}{2}=0\) for all integers \(n\), we find \[\begin{align*} A_n &= \frac{1}{2}e^{-\frac{1}{2}(4n+1)\pi}\left(1 + 0 -2 \right) - \frac{1}{2}e^{-\frac{1}{2}(4n+1)\pi}e^{2\pi}\left(1 + 0 -2\right) \\ &= \frac{1}{2}e^{-\frac{1}{2}(4n+1)\pi}\left(-1 +e^{2\pi} \right) \\ &= \frac{1}{2}\left(e^{2\pi} -1\right)e^{-\frac{1}{2}(4n+1)\pi}, \end{align*}\]

as required.

[…] and hence find \(\displaystyle\sum\limits_{n=1}^{\infty} A_n\).

Let \(\,A=\displaystyle\sum\limits_{n=1}^{\infty} A_n\). Then \[\begin{align*} A &= \sum\limits_{n=1}^{\infty} \frac{1}{2}\left(e^{2\pi} -1\right)e^{-\frac{1}{2}(4n+1)\pi} \\ A &= \frac{1}{2}\left(e^{2\pi} -1\right) \sum\limits_{n=1}^{\infty} e^{-2n\pi}e^{-\frac{\pi}{2}} \\ A &= \frac{1}{2}\left(e^{2\pi} -1\right) e^{-\frac{\pi}{2}} \sum\limits_{n=1}^{\infty} (e^{-2\pi})^n \\ A &= \frac{1}{2}\left(e^{2\pi} -1\right) e^{-\frac{\pi}{2}} \left[e^{-2\pi} + (e^{-2\pi})^2 + (e^{-2\pi})^3 + \dotsb\right]\\ A &= \frac{1}{2}\left(e^{2\pi} -1\right) e^{-\frac{\pi}{2}} e^{-2\pi} \left[1+ e^{-2\pi} + (e^{-2\pi})^2 + \dotsb\right] \end{align*}\] We therefore have the infinite sum of a geometric series, with first term \(\,a=\frac{1}{2}\left(e^{2\pi} -1\right) e^{-\frac{5\pi}{2}}\) and ratio \(\,r=e^{-2\pi}\). Since \(\,2\pi > 0\), \(e^{-2\pi}<1\) and the series converges, with infinite sum \(\,S_{\infty}=\frac{a}{1-r}\). Therefore \[\begin{align*} A &= \frac{1}{2}\left(e^{2\pi} -1\right) e^{-\frac{5\pi}{2}}\times \frac{1}{1-e^{-2\pi}} \\ &= \frac{1}{2}\left(e^{2\pi} -1\right) e^{-\frac{5\pi}{2}} \times \frac{e^{2\pi}}{e^{2\pi}-1} \\ &= \frac{1}{2}e^{-\frac{5\pi}{2}} \times e^{2\pi} = \frac{1}{2}e^{-\frac{\pi}{2}}. \end{align*}\]

So \(\displaystyle\sum\limits_{n=1}^{\infty} A_n = \frac{1}{2}e^{-\frac{\pi}{2}}\).